Solitaire has N elevators. Each elevator are connecting exactly two floors and it does not stop on the floors between that two floors The speed of all the elevators are the same, 5 seconds to pass one floor.

On the beginning, each elevator is in its lower position and they are starting cruising to the upper floor. After some elevator come to its upper position, it immediatly starts to go back to its lower position, and so on...

Mirko is on the first (the lowest) floor and he wants as quick as possible come to the top of the solitaire. He can change elevators only on the floors that are common to both elevators, and if the other elevator is in that moment on that floor, that change does not take any time.

Write a program that will calculate minimal time in which Mirko can get to the top of the solitaire.


In the first line of the input file there are two integers K and N, separated with space, number of floors in solitaire and number of elevators, 2 ≤ K ≤ 1000, 1 ≤ N ≤ 50000.

In each of the next N lines there are description of one elevator, two integers A and B, separated with space, 1 ≤ A < B ≤ K, means that elevator is travelling between floors A and B.

There are no two different elevators that travels between same floors.

Note: input data will guarantee that solution will always exists.


In the only line of output file write minimal time (in seconds) from the text above.

10 4 
1 5 
5 10 
5 7 
7 10 
10 3 
1 5 
3 5 
3 10 
20 5 
1 7 
7 20 
4 7 
4 10 
10 20 

hide comments
manishjoshi394: 2018-12-10 09:31:59

Just create a function that will calculate weights of the edges depending on the time already elapsed. Simple dijkstra.

sherlock11: 2018-05-25 06:49:10

toughest part of question is to understand the question..........after reaching to a floor he can wait on that floor,the elevator don't need to be synchronous at the time for mirko to change the elevator.......this misunderstanding costed me 3 days and and raised question on my existence...finally i have answers for all the the way dijkstra with function to calculate wait time and there u go AC:)

Last edit: 2018-05-25 06:49:59
badry atef: 2016-06-03 15:49:29

normal dijkstra's algorithm with binary search to calculate waiting time got it ACC :D

xamitksx: 2016-04-08 19:27:13

NO STL is required . . variant of BFS ...

marky: 2016-03-02 18:33:22

How to calculate wait time for next lift?

Russo: 2015-11-09 00:40:26

--DO NOT-- read/write from/to the files indicated. Just ignore those lines of the input and read from stdin and write to stdout.

Ankur Singh: 2015-06-22 08:02:56

can he change the elevator even when the elevator does not have a stoppage(begin / end) point. Like if one elevator goes from 1 - 5, other 3 - 5, can he change at floor 4 ?

rush: 2015-03-19 18:42:59

Nice variation of a classical algorithm.

Amitayush Thakur: 2014-09-02 18:05:32

My 100th submission on SPOJ :)

Ninjaflyte: 2010-06-17 15:48:00

@~!(*(@*!@^& : I am completely sure that my program is correct, yet its getting WA. Can you check what's wrong?
(or at least give a test case where it fails)
Submission ID : 3740920

Added by:~!(*(@*!@^&
Time limit:1s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
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Resource:COI 03